ENERGY BALANCE

ENERGY BALANCE ON VAPORIZER:

First of all I will calculate the heat duty of vaporizer:

Q= mCp∆T + mλ

= 429.4*1.9228*77.5+429.4*0.446                                           T2 = 58oC

= 64179.4224kj/hr

Reference temperature = 25°C

Heat energy IN = heat energy OUT

Heat energy in = heat in with SBA + heat in with steam

Heat in with SBA

Heat energy in = mCpDT                                            T = 30oC

Mass of SBA in = 429.4 kg/hr

Cp of liquid SBA at 30°C (86°F)

Cp = 0.48 (Btu/lb°F)

(REF .FIG # 4 KERN)                       T1 = 140oC

AS,  (Btu/lb°F) = (Cal/gm°C)

Cp=

0.48 4.18 joule 1KJ T = 140°C

1000 gm

       Gm °C 1 1000 J 1 Kg

—Cp = 2.0 KJ/Kg°C

—  DT = (30-25)°C

—        = 5°C

—Heat energy in = mCpDT

429.4Kg 2.00KJ 5°C
hr Kg°C

—Heat in with SBA =  4294kj/hr

—

—Energy in with steam

— Energy in with steam = mCp DT

—Cp of steam at 160o C = 1.168 KJ/kg. K

—

—

—

—Energy IN with steam=2733(ms)

—

—Heat energy IN=heat in with SBA + heat in with steam

—  =17009+2733(ms)

—Energy out

—

—Energy out by vap. SBA

— 

—Energy out by vap. SBA =mCpDT

—

—Mass of SBA out=429.4kg/hr

—Cp of liquid SBA at 107.5° (225.5°F)

—  (REF. FIG # 5,KERN)

—

—As  (Btu/lb°F)=(Cal/gm °C)

0.44 4.18 joule 1 KJ 1000 gm
Gm °C 1 1000 J 1 Kg

—Cp = 1.8392KJ/kg. o  C

—Energy out by vap. SBA = mCpDT

429.4kg 1.83KJ 77.5°C
hr Kg°C

= 61205.8 KJ/hr

—Heat energy out with steam:

— Assume the outlet temperature of the steam is 58°C (136°F)

—         = 2605.9ms

—

—Total heat energy out = Heat out with SBA + Heat out with steam

—Heat energy out         = 61205.81+ 2605.9ms

—

—Heat energy IN = heat energy OUT

—

—4294+ 2733ms = 61205.81+ 2605.9ms

—127.1ms  =  56911.81

—ms = 444.6215Kg/hr

Energy balance on reactor

—As described in the process, feed out of the reactor is at temp of 340°C

—

—

—Reference temp = 25°C

—

—Heat energy IN = heat energy OUT

—

—As reaction is endothermic

—So.

—Heat energy IN = heat energy OUT + Total heat of reaction

—

—Heat energy out

—Heat energy out = mCpDT

—Mass of product out of reactor = 471Kg/hr

—

—As we know product out of reactor is a mixture of SBA, MEK, H2, so total heat capacity is the heat capacity of mixture.

—

—Heat capacity at 340°C (644°F)

—

—Cp of MEK

—Cp MEK = 0.67 Btu/lb°F

—  (REF. FIG # 5, KERN)

—

—AS,  (Btu/lb°F) = (cal/gm°C)

—Cp MEK =

0.67 4.18 J 1000 gm 1 KJ
gm°C 1 Kg 1000J

—Cp MEK = 2.8 KJ/Kg°C

—

—% of MEK at reactor = 86.78 %

—Cp MEK = 0.8678 * 2.80

—     = 2.43 KJ/Kg°C

—

—Cp of SBA

—Cp SBA = 0.65 Btu/lb°F

—

—AS,   (Btu/lb°F) = (Cal/gm°C)

—

—Cp SBA =

0.65 4.18 J 1000 gm 1KJ
gm°C 1 Kg 1000 J

—Cp SBA = 2.71 KJ/Kg°C

—

—% of SBA out of reactor = 10.5%

—

—Cp SBA = 2.71 * 0.105

—     = 0.2614 KJ/Kg°C

—Cp of H2 = 3.5 Btu/lb°F

—  (REF, FIG# 3, D.Q KERN)

—AS,   (Btu/lb°F) = (Cal/gm°C)

—

—Cp H2 =

3.5 4.18 J 1000 gm 1KJ
gm°C 1 Kg 1000 J

—Cp H2 = 14.63 KJ/Kg°C

—% of H2 out of reactor = 2.73 %

—Cp H2 = 14.6363 * 0.0274

—    = 0.4068 KJ/Kg°C

—

—Cp mix = Cp MEK + Cp SBA + Cp H2

—    = 2.43 + 0.2614 + 0.4068

—    = 3.1226 KJ/Kg°C

—

—Heat energy out = mCpDT

—

—Heat energy out = = 4.63* 10­­6 KJ/hr

—Total heat of reaction

—

—Heat of reaction (at mean reaction temp) = 560 KJ/kg mole

—

—Mol. Wt of SBA = 74 KJ/kg mole

—From stoichiometry SBA required = 471 Kg/hr

—Total heat of reaction =

560 KJ Kg mole 471 kg
Kg mole 74 kg Hr

—= 1.2* 104 KJ/hr

—

—Heat out = 4.63 * 10­­6 + 1.2 * 104

—Heat out = 4.64*106 KJ/hr

—

—Heat energy in

—

—Heat energy in = mCpDT

—

—As reaction is endothermic so heat is required, suppose an inlet temperature 550°C (1022°F)

—Inlet feed has a mixture of SBA & MEK so Cp is Cp mix.

—

—Cp mix = Cp MEK + Cp SBA

—

—Cp SBA

—Cp SBA = 0.8 Btu/lb°F

—

—AS,   (Btu/lb°F) = (Cal/gm°C)

—Cp SBA =

0.80 4.18 J 1000 gm 1KJ
gm°C 1 Kg 1000 J

—Cp SBA = 3.43 KJ/kg°C

—

—% of SBA in feed = 99.99%

—Cp SBA = 3.43 * 0.9999

—     = 3.42 KJ/kg°C

—

—Cp MEK

—Cp MEK = 0.82 Btu/lb°C

—

—AS,   (Btu/lb°F) = (Cal/gm°C)

—Cp MEK =

0.82 4.18 J 1000 gm 1KJ
gm°C 1 Kg 1000J

—Cp MEK = 3.42 KJ/kg°C

—

—% of MEK in feed = 0.01 %

—Cp MEK = 3.43*0.0001

—     = 3.42*10-4 KJ/kg°C

—Cp mix = Cp MEK + Cp SBA

—    = 3.42+3.42*10-4

—    = 3.42 KJ/kg°C

—Heat energy in = mCpDT

—         = 1470.96*3.42*(T-25)

—

—Heat energy In = heat energy OUT

—

— 470.96*3.42*(T-25) = 4.64*106

—

—  T = 545°C

—

—So, it is near to our assumption so take inlet temperature 429°C.

—So, feed inlet temperature is 420°C.

 

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